「力扣」周赛第 170 周题解(Java)

这周做出来 2 题。

  • 第 1 题调试了很久,后来才发现,原来只需要一个 ifelse 就解决了;
  • 第 2 题暴力做的,没有想到异或也适用于前缀和;
  • 第 3 题就是基础不好了,是几个基础问题的综合题;
  • 第 4 题这周刚刚总结过回文子串和动态规划的技巧,中午趴在桌子上睡一觉,居然还就找出原因了,打印出 dp 数组看了一眼,就发现状态转移出错了。

感谢朋友“大肥凯”的鼓励,做了几周周赛,几乎困难题都是放弃的,哈哈哈哈。这周困难题不一样了,事后自己做了出来,还是有进步哦。

第 1 题:解码字母到整数映射

竞赛页地址:https://leetcode-cn.com/problems/decrypt-string-from-alphabet-to-integer-mapping/

Java 代码:

public class Solution {

    public String freqAlphabets(String s) {
        StringBuilder stringBuilder = new StringBuilder();
        int len = s.length();

        char[] charArr = new char[]{' ', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i'};

        for (int i = 0; i < len; i++) {
            if (i + 2 < len && s.charAt(i + 2) == '#') {
                int index = Integer.parseInt(s.substring(i, i + 2));
                stringBuilder.append((char) (index + 96));
                i += 2;
            } else {
                stringBuilder.append(charArr[s.charAt(i) - '0']);
            }
        }
        return stringBuilder.toString();
    }
}

第 2 题:子数组异或查询

竞赛页地址:https://leetcode-cn.com/problems/xor-queries-of-a-subarray/

方法一:直接计算

Java 代码:

import java.util.Arrays;

public class Solution {

    public int[] xorQueries(int[] arr, int[][] queries) {
        int qLen = queries.length;
        int[] res = new int[qLen];

        for (int i = 0; i < qLen; i++) {
            int leftIndex = queries[i][0];
            int rightIndex = queries[i][1];

            int curRes = 0;
            for (int j = leftIndex; j <= rightIndex; j++) {
                curRes ^= arr[j];
            }
            res[i] = curRes;
        }
        return res;
    }
}

方法二:依然可以使用前缀和的技巧。

Java 代码:

import java.util.Arrays;

public class Solution {

    public int[] xorQueries(int[] arr, int[][] queries) {
        int len = arr.length;

        // 得到前缀异或和数组
        int[] preSum = new int[len + 1];
        for (int i = 0; i < len; i++) {
            preSum[i + 1] = preSum[i] ^ arr[i];
        }

        int qLen = queries.length;
        int[] res = new int[qLen];

        for (int i = 0; i < qLen; i++) {
            int left = queries[i][0];
            int right = queries[i][1];
            // 注意:这里也是取异或
            res[i] = preSum[right + 1] ^ preSum[left];
        }
        return res;
    }
}

第 3 题:获取你好友已观看的视频

竞赛页地址:https://leetcode-cn.com/contest/weekly-contest-170/problems/minimum-insertion-steps-to-make-a-string-palindrome/

Java 代码:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Queue;

public class Solution {

    public List<String> watchedVideosByFriends(List<List<String>> watchedVideos, int[][] friends, int id, int level) {
        int len = friends.length;

        // 距离 id 的步数,借助队列,初值为 -1 表示还没有赋值
        int[] distance = new int[len];
        Arrays.fill(distance, -1);

        distance[id] = 0;
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(id);

        while (!queue.isEmpty()) {
            Integer top = queue.poll();

            for (int friend : friends[top]) {
                if (distance[friend] == -1) {
                    distance[friend] = distance[top] + 1;
                    queue.offer(friend);
                }
            }
        }

        List<String> res = new ArrayList<>();

        // 以下同「力扣」第 451 题:根据字符出现频率排序

        // 存储了距离为 level 的朋友所看的电影和对应的观看次数
        Map<String, Integer> vatchedTimes = new HashMap<>();

        for (int i = 0; i < len; i++) {
            // 只关心距离为 level 的朋友所看的电影
            if (distance[i] == level) {
                for (String video : watchedVideos.get(i)) {
                    if (vatchedTimes.get(video) == null) {
                        res.add(video);
                        vatchedTimes.put(video, 0);
                    } else {
                        vatchedTimes.put(video, vatchedTimes.get(video) + 1);
                    }
                }
            }
        }

        res.sort((v1, v2) -> {
            if (!vatchedTimes.get(v1).equals(vatchedTimes.get(v2))) {
                return vatchedTimes.get(v1) - vatchedTimes.get(v2);
            }

            return v1.compareTo(v2);
        });
        return res;
    }

    public static void main(String[] args) {
        List<List<String>> watchedVideos = new ArrayList<>();

        List<String> video1 = new ArrayList<>();
        video1.add("A");
        video1.add("B");

        List<String> video2 = new ArrayList<>();
        video2.add("C");


        List<String> video3 = new ArrayList<>();
        video3.add("B");
        video3.add("C");

        List<String> video4 = new ArrayList<>();
        video4.add("D");


        watchedVideos.add(video1);
        watchedVideos.add(video2);
        watchedVideos.add(video3);
        watchedVideos.add(video4);


        int[][] friends = {{1, 2}, {0, 3}, {0, 3}, {1, 2}};

        int id = 0;
        int level = 2;

        Solution solution = new Solution();
        List<String> res = solution.watchedVideosByFriends(watchedVideos, friends, id, level);

        System.out.println(res);
    }
}

第 4 题:让字符串成为回文串的最少插入次数

竞赛页地址:https://leetcode-cn.com/problems/minimum-insertion-steps-to-make-a-string-palindrome/

心得:状态转移,即让程序“填表”的时候,方向一定要搞对!

Java 代码:

import java.util.Arrays;

public class Solution {

    public int minInsertions(String s) {
        int len = s.length();

        int[][] dp = new int[len][len];
        for (int j = 1; j < len; j++) {
            
            // 注意顺序,之后的值一定会被之前参考到
            for (int i = j - 1; i >= 0; i--) {
                if (s.charAt(i) == s.charAt(j)) {
                    if (j - i < 3) {
                        dp[i][j] = 0;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                } else {
                    if (j - i < 2) {
                        dp[i][j] = 1;
                    } else {
                        dp[i][j] = Math.min(dp[i + 1][j], dp[i][j - 1]) + 1;
                    }
                }
            }
        }

//        for (int i = 0; i < len; i++) {
//            System.out.println(Arrays.toString(dp[i]));
//        }
        return dp[0][len - 1];
    }
}

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